Results 1 to 6 of 6

Math Help - another identity

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    19

    another identity

    Prove...
    \frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)

    this is as far as i got
    working from LHS

    \frac{1}{sec(x)-tan(x)}

    cos(x)-cot(x)

    cos(x)-\frac{cos(x)}{sin(x)}

    \frac{sin(x)}{sin(x)}cos(x)-\frac{cos(x)}{sin(x)}

    \frac{sin(x)cos(x)}{sin(x)}-\frac{cos(x)}{sin(x)}

    now i'm stuck..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    You must know that (a-b)(a+b)=a^2-b^2. This identity is worth being very familiar with.

    So you want to prove 1=\sec^2x-\tan^2x (I just multiplied both sides of your identity by \sec x - \tan x).
    Now just apply the identity \sec^2x=1+\tan^2x and you're done!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by windir View Post
    Prove...
    \frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)

    this is as far as i got
    working from LHS

    \frac{1}{sec(x)-tan(x)}

    cos(x)-cot(x)

    cos(x)-\frac{cos(x)}{sin(x)}

    \frac{sin(x)}{sin(x)}cos(x)-\frac{cos(x)}{sin(x)}

    \frac{sin(x)cos(x)}{sin(x)}-\frac{cos(x)}{sin(x)}

    now i'm stuck..
    \frac{1}{\sec x-\tan x}{\color{red}\neq}\frac{1}{\sec x}-\frac{1}{\tan x}!!!! XD

    What I suggest you do with the LHS is the following: \frac{1}{\sec x-\tan x}=\frac{1}{\sec x-\tan x}\cdot\frac{\sec x+\tan x}{\sec x+\tan x}=\dots

    In the process, invoke one of the pythagorean identities...

    Can you take it from here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
     \frac{1}{\sec x - \tan x} \cdot \frac{\sec x + \tan x}{\sec x + \tan x}

     = \frac {\sec x + \tan x}{\sec^{2} x - \tan^{2} x }

     = \frac {\sec x + \tan x}{1 + \tan^{2}x - \tan^{2}x}

     = \sec x + \tan x
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2009
    Posts
    19
    ooooooooooohhh, duh! thanks man!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by windir View Post
    Prove...
    \frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)

    [snip]
    Also discussed here: http://www.mathhelpforum.com/math-he...589-proof.html

    Thread closed.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. the i*pi=ln(-1) identity
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: August 1st 2010, 04:26 AM
  2. identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 24th 2010, 06:50 AM
  3. Identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 11th 2010, 08:32 AM
  4. An identity
    Posted in the Math Challenge Problems Forum
    Replies: 0
    Last Post: April 7th 2009, 10:22 PM
  5. identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 21st 2009, 09:40 PM

Search Tags


/mathhelpforum @mathhelpforum